# Thread: Multiplicity of a root

1. ## Multiplicity of a root

Let c be a fixed real number.
Show that a root of the equation
x(x+1)(x+2)...(x+2009)=c
can have multiplicity at most 2. Determine the number of values of c for which the equation has a root of multiplicity 2.

Can anybody tell me how to compute this?

2. Originally Posted by nahduma
Let c be a fixed real number.
Show that a root of the equation
x(x+1)(x+2)...(x+2009)=c
can have multiplicity at most 2. Determine the number of values of c for which the equation has a root of multiplicity 2.
If an equation f(x) = c has a root x=a of multiplicity 2 then f'(a) = 0 (I know this is the pre-calculus forum, but I don't think you'll be able to do this question without using differentiation). If the root has multiplicity greater than 2, then also f"(a) = 0.

For the equation x(x+1)(x+2)...(x+2009) = c, take logs to get $\displaystyle \sum_{n=0}^{2009}\ln(x+n) = \ln c$. Then let $\displaystyle f(x) = \sum_{n=0}^{2009}\ln(x+n)$. Differentiate to get $\displaystyle f'(x) = \sum_{n=0}^{2009}\frac1{x+n}$. Then differentiate again to get $\displaystyle f''(x) = -\sum_{n=0}^{2009}\frac1{(x+n)^2}$. Since all the terms in that last sum have the same sign, it's obvious that their sum cannot be zero. So $\displaystyle f(x) = \ln c$ has no roots of multiplicity greater than 2.

To find the number of roots of multiplicity 2, notice that the derivative of a function has a root between each pair of consecutive roots of the function. Since x(x+1)(x+2)...(x+2009) = 0 has 2010 roots, its derivative will have 2009 roots.